The reaction 2A = B is the second order reaction in A with a rate constant of 51/M × min at 24°C.
1. Starting with (A) initial=0.0092M, how long will it take in seconds for (A) final= 3.7 × 10^-3M
2. Calculate the half-life of the reaction?
1. To calculate the time it takes for A to decrease from an initial concentration of 0.0092M to a final concentration of 3.7 x 10^-3M, we can use the integrated rate equation for a second-order reaction:
1/[A]t - 1/[A]0 = kt
where:
[A]t = final concentration of A = 3.7 x 10^-3 M
[A]0 = initial concentration of A = 0.0092 M
k = rate constant = 51/M x min
Plugging in the values, we get:
1/3.7 x 10^-3 - 1/0.0092 = 51t
Solving for t:
t ≈ 25.81 min ≈ 1549 seconds
So, it will take approximately 1549 seconds for A to decrease from 0.0092M to 3.7 x 10^-3M.
2. To calculate the half-life of the reaction, we can use the equation:
t1/2 = 1 / (k[A]0)
Plugging in the values:
t1/2 = 1 / (51 x 0.0092) = 0.022 M / min
Therefore, the half-life of the reaction is approximately 0.022 minutes.